2022 AMC 10B 第 6 题

Question

数列

121, 11211, 1112111, \ldots

的前十项中有多少个质数？

How many of the first ten numbers of the sequence

121, 11211, 1112111, \ldots

are prime numbers?

0

1

2

3

4

把第

n

项表示成两个重叠的、各由

n+1

个一组成的数之和

Express the

n

th term as the sum of two overlapping blocks of

n+1

ones

每一项都可以分解为两个大于一的因数。

Each term in the sequence has an obvious factorization pattern

我们断言这些数都不可能是质数。

第

n

项可写成

\sum_{k=0}^{2n} 10^k + 10^n = \sum_{k=0}^{n} 10^k +

\sum_{k=n}^{2n} 10^k =\sum_{k=0}^{n} 10^k + \sum_{k=0}^{n} 10^k \cdot 10^n

= (10^n+1)(\sum_{k=0}^{n} 10^k)\text{。}

这说明每一项都能写成两个大于

1

的整数的乘积，所以前十项中没有质数。

所以正确答案是 A。

We claim that none of these numbers can ever be prime.

We prove this claim by noticing that the

n

th number is

\sum_{k=0}^{2n} 10^k + 10^n = \sum_{k=0}^{n} 10^k +

\sum_{k=n}^{2n} 10^k =\sum_{k=0}^{n} 10^k + \sum_{k=0}^{n} 10^k \cdot 10^n

= (10^n+1)(\sum_{k=0}^{n} 10^k).

This shows that the number can be written as the product of two numbers greater than

1,

so there are no primes.

Thus, the answer is A .

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